Consider the first-order logic sentence $F: \forall x(\exists y R(x, y))$. Assuming non-empty logical domains, which of the sentences below are *implied* by $F$? I. $\exists y(\exists x R(x, y))$ II. $\exists y(\forall x R(x, y))$ III. $\forall y(\exists x R(x, y))$ IV. $eg \exists x(\forall y eg R(x, y))$

We are given the sentence $F: \forall x (\exists y R(x, y))$ over a non-empty domain $D$.

Let us evaluate each statement:

• Statement I: $\exists y (\exists x R(x, y)) \equiv \exists x \exists y R(x, y)$. Since $D$ is non-empty, there exists at least one element $c \in D$. From $F$, since $\forall x (\exists y R(x, y))$ holds, it must hold for $c$, so $\exists y R(c, y)$ is true. Hence, $\exists x \exists y R(x, y)$ is true. Thus, $F \implies \text{I}$.

• Statement II: $\exists y (\forall x R(x, y))$. $F$ states that every $x$ has some related $y$, but not necessarily the *same* $y$ for all $x$. Hence, $\forall x \exists y R(x, y)

ot\implies \exists y \forall x R(x, y)$.

• Statement III: $\forall y (\exists x R(x, y))$. $F$ does not guarantee that every $y$ is related to some $x$. For instance, on the domain of natural numbers $\mathbb{N}$ where $R(x, y)$ means $x < y$, for every $x$ there exists a $y > x$, but for $y = 0$, there is no $x < 0$. Thus, $F

ot\implies $\text{III}$$.

• Statement IV: $

eg \exists x (\forall yeg R(x, y))$.Using quantifier negation laws: $$eg \exists x (\forall y eg R(x, y)) \equiv \forall x eg (\forall y eg R(x, y)) \equiv \forall x (\exists y eg eg R(x, y)) \equiv \forall x (\exists y R(x, y)) = F$$ Since IV is logically equivalent to $F$, it is trivially implied by $F$.

Thus, statements I and IV only are implied by $F$, matching option B.